How do you find the lengths of the curve $8 x = 2 y^{4} + y^{- 2}$ $8x=2y^4+y^-2$ for $1 \leq y \leq 2$ $1<=y<=2$ ?

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Answer 1 · 2017-04-26T08:11:21

$8 x = 2 y^{4} + y^{- 2}$ $8x=2y^4+y^(-2)$

$implies x'=y^3- 1/4 y^(-3)$

$s = int_1^2 sqrt( 1 + (x')^2) \ dy$

$= int_1^2 sqrt( 1 + (y^3- 1/4 y^(-3))^2) \ dy$

$= int_1^2 sqrt( 1 + y^6 + 1/16 y^(-6)- 1/2) \ dy$

$= int_1^2 sqrt( y^6 + 1/16 y^(-6) + 1/2) \ dy$

$= int_1^2 sqrt( (y^3 + 1/4 y^(-3))^2) \ dy$

$= int_1^2 y^3 + 1/4 y^(-3) \ dy$

$= [ y^4/4 - 1/(8 y^2) ]_1^2$

$= 123/32$

How do you find the lengths of the curve 8x=2y^4+y^-28x=2y4+y−28x=2y^4+y^-2 for 1<=y<=21≤y≤21<=y<=2?